Question

用数学归纳法证明

(1×2²-2×3²)+(3×4²-4×5²)+…+［(2n-1)(2n)²-2n(2n+1)²］=-n(n+1)(4n+3)(n∈N^*).

Answer 1

证明:(1)当n=1时,左边=1×2²-2×3²=-14,右边=-1×2×7=-14,等式成立.?

(2)假设当n=k时等式成立,即

(1×2²-2×3²)+(3×4²-4×5²)+…+［(2k-1)(2k)²-2k(2k+1)²］=-k(k+1)(4k+3),则当n=k+1时,

(1×2²-2×3²)+(3×4²-4×5²)+…+［(2k-1)(2k)²-2k(2k+1)²］+［(2k+1)(2k+2)²-(2k+2)(2k+3)²］

=-k(k+1)(4k+3)-2(k+1)［(4k²+12k+9)-(4k²+6k+2)］?

=-k(k+1)(4k+3)-2(k+1)(6k+7)

=-(k+1)(4k²+15k+14)?

=-(k+1)(k+2)(4k+7)?

=-(k+1)［(k+1)+1］［4(k+1)+3］.?

这说明当n=k+1时,等式也成立.?

由(1),(2)可知等式对n∈N^*都成立.

温馨提示

用数学归纳法证明恒等式,关键是在证明n=k+1时命题成立.从n=k+1时的待证恒等式的一端“拼凑”出归纳假设恒等式的一端，再运用归纳假设即可.